Lesson: Triangles - Part II

Comment on Triangles - Part II

At 7:21, if we draw an altitude to the isosceles triangle with one of the equal sides as base, would the altitude still bisect it?
gmat-admin's picture

Good question.
The answer is no. In order for the diagonal to bisect the top angle, the non-equal side must be the base.
Try it with a 45-45-90 triangle, and you'll see what I mean.

Hi Brent,

A little confused with this. Suppose if the isosceles triangle is ABC with sides AB=AC, then will the altitude drawn from angle A bisect angle BAC in two equal parts? I thought it is only a bisector segment dividing side BC in two equal parts.

Thanks
gmat-admin's picture

Yes, the altitude drawn from point A WILL bisect angle BAC in two equal parts.

Think of it this way (using the diagram you describe with point D representing the point where the altitude intersects side BC)

We get two triangles: ∆ADB and ∆ADC

First, recognize that ∠ABD = ∠ACD, since those are the equal angles in ∆ABC

Second, recognize that ∠ADB = ∠ADC = 90°, since AD is the ALTITUDE

Third, since triangles ∆ADB and ∆ADC already have two angles in common, the third angles must also be equal. That is ∠BAD = ∠CAD

Does that help?

Cheers,
Brent

Amazing. Without an inkling of doubt.
Is a bisector drawn always perpendicular to the line it bisects or this is the property applicable only in case of Isosceles and Equilateral triangle?

Thanks.
gmat-admin's picture

The bisector is perpendicular ONLY IF the triangle is either isosceles and equilateral.

Sir is there any easy solution for the question given below
https://gmatclub.com/forum/in-the-figure-above-point-o-is-the-center-of-the-circle-and-107874.html
gmat-admin's picture

Have you seen Karishma's solution: https://gmatclub.com/forum/in-the-figure-above-point-o-is-the-center-of-...

If you have any questions about her solution, let me know and we can go from there.

sir, i am having doubt in this also,

https://gmatclub.com/forum/in-triangle-abc-point-x-is-the-midpoint-of-side-ac-and-82671.html
gmat-admin's picture

One of the statements from the DS question reads as follows
"(1) x^2 + y^2≠ z^2"

Does this statement translate into the triangle being an equilateral triangle?
gmat-admin's picture

Question link: https://gmatclub.com/forum/is-the-area-of-the-triangular-region-above-le...

IF it were the case that x² + y² = z², then we'd know that the triangle is an RIGHT TRIANGLE with z as the hypotenuse.

Since we're told that x² + y² ≠ z², then we know that z is not the hypotenuse of a right triangle.

ASIDE #1: This does not necessarily mean that the given triangle is not a right triangle. It could be the case that side x is the hypotenuse of the triangle, in which case, we'd get y² + z² = x²

ASIDE #2: It COULD also be the case that the given triangle is an equilateral triangle, but it could also not be an equilateral triangle.

Does that help?

Cheers,
Brent

Hi Brent

Helps perfectly thank you, I saw in a later video this topic is discussed in more depth.

Regards

Hi Brent,

Could you please present your answer on this?
https://gmatclub.com/forum/in-triangle-abc-point-x-is-the-midpoint-of-side-ac-and-82671.html

The question introduces properties of Midsegements, something i haven't come across in our course module.

Thanks.
gmat-admin's picture

You bet!

Here's my step-by-step solution: https://gmatclub.com/forum/in-triangle-abc-point-x-is-the-midpoint-of-si...

Cheers,
Brent

Question Link: https://gmatclub.com/forum/in-the-figure-above-point-o-is-the-center-of-the-circle-and-107874.html

Would you mind explaining the solution. Specifically I would like to know the approach on how the deduce or crack the given info. I seem to understand but it when encountered again, I can't get it right
gmat-admin's picture

Hi emailme,

Here's my step-by-step solution: https://gmatclub.com/forum/in-the-figure-above-point-o-is-the-center-of-...

Cheers,
Brent

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